JEE Main On 8 April 2017 Question 10
Question: A uniform disc of radius R and mass M is free to rotate only about its axis. A string is wrapped over its rim and a body of mass m is tied to the free end of the string as shown in the figure. The body is released from rest. Then the acceleration of the body is - [JEE Online 08-04-2017]
Options:
A) $ \frac{2Mg}{2m+M} $
B) $ \frac{2Mg}{2M+m} $
C) $ \frac{2mg}{2M+m} $
D) $ \frac{2mg}{2m+M} $
Show Answer
Answer:
Correct Answer: D
Solution:
- mg - T = ma $ RT=I\propto $ $ RT=\frac{MR^{2}}{2}.\frac{a}{R} $ $ T=\frac{Ma}{2} $ $ mg-\frac{Ma}{2}=ma $ $ mg=a( \frac{M}{2}+m ) $ $ mg=a( \frac{M+2m}{2} ) $ $ a=\frac{2mg}{M+2m} $
